Laplace Transform of Identity Mapping
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Theorem
Let $\laptrans f$ denote the Laplace transform of a function $f$.
Let $\map {I_\R} t$ denote the identity mapping on $\R$ for $t > 0$.
Then:
- $\laptrans {\map {I_\R} t} = \dfrac 1 {s^2}$
for $\map \Re s > 0$.
Proof 1
\(\ds \laptrans {\map {I_\R} t}\) | \(=\) | \(\ds \laptrans t\) | Definition of Identity Mapping | |||||||||||
\(\ds \) | \(=\) | \(\ds \int_0^{\to +\infty} t e^{-s t} \rd t\) | Definition of Laplace Transform | |||||||||||
\(\ds \) | \(=\) | \(\ds \intlimits {\frac {e^{-s t} } {-s} \paren {t - \frac 1 {-s} } } {t \mathop = 0} {t \mathop \to +\infty}\) | Primitive of $x e^{a x}$ | |||||||||||
\(\ds \) | \(=\) | \(\ds -\frac 1 s \lim_{t \mathop \to +\infty} \frac t { e^{s t} } - \paren {0 - \frac 1 {s^2} }\) | Exponential of Zero and One, Exponent Combination Laws: Negative Power | |||||||||||
\(\ds \) | \(=\) | \(\ds 0 + \frac 1 {s^2}\) | Limit at Infinity of Polynomial over Complex Exponential | |||||||||||
\(\ds \) | \(=\) | \(\ds \frac 1 {s^2}\) |
$\blacksquare$
Proof 2
\(\ds \laptrans {\map {I_\R} t}\) | \(=\) | \(\ds \laptrans t\) | Definition of Identity Mapping | |||||||||||
\(\ds \) | \(=\) | \(\ds \int_0^{\to +\infty} t e^{-st} \rd t\) | Definition of Laplace Transform |
From Integration by Parts:
- $\ds \int f g' \rd t = f g - \int f'g \rd t$
Here:
\(\ds f\) | \(=\) | \(\ds t\) | ||||||||||||
\(\ds \leadsto \ \ \) | \(\ds f'\) | \(=\) | \(\ds 1\) | Derivative of Identity Function | ||||||||||
\(\ds g'\) | \(=\) | \(\ds e^{-st}\) | ||||||||||||
\(\ds \leadsto \ \ \) | \(\ds g\) | \(=\) | \(\ds -\frac 1 s e^{-s t}\) | Primitive of Exponential Function |
So:
\(\ds \int t e^{-s t} \rd t\) | \(=\) | \(\ds -\frac t s e^{-s t} - \frac 1 s \int e^{-s t} \rd t\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds -\frac t s e^{-s t} - \frac 1 {s^2} e^{-s t}\) | Primitive of Exponential Function |
Evaluating at $t = 0$ and $t \to +\infty$:
\(\ds \laptrans t\) | \(=\) | \(\ds \intlimits {-\frac t s e^{-s t} - \frac 1 {s^2} e^{-s t} } {t \mathop = 0} {t \mathop \to +\infty}\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds -\frac 1 s \lim_{t \mathop \to +\infty} \frac t { e^{s t} } - \paren {0 - \frac 1 {s^2} }\) | Exponential of Zero and One, Exponent Combination Laws: Negative Power | |||||||||||
\(\ds \) | \(=\) | \(\ds 0 + \frac 1 {s^2}\) | Limit at Infinity of Polynomial over Complex Exponential | |||||||||||
\(\ds \) | \(=\) | \(\ds \frac 1 {s^2}\) |
$\blacksquare$
Proof 3
From Laplace Transform of Derivative:
- $(1): \quad \laptrans {\map {I_\R'} t} = s \laptrans {\map {I_\R} t} - \map {I_\R} 0$
under suitable conditions.
Then:
\(\ds \map {I_\R} t\) | \(=\) | \(\ds t\) | ||||||||||||
\(\ds \leadsto \ \ \) | \(\ds \map {I_\R'} t\) | \(=\) | \(\ds 1\) | |||||||||||
\(\ds \map {I_\R} 0\) | \(=\) | \(\ds 0\) | ||||||||||||
\(\ds \leadsto \ \ \) | \(\ds \laptrans 1\) | \(=\) | \(\ds \dfrac 1 s\) | Laplace Transform of 1 | ||||||||||
\(\ds \) | \(=\) | \(\ds s \laptrans {\map {I_\R} t} - 0\) | from $(1)$, substituting for $\map f t$ and $\map f 0$ | |||||||||||
\(\ds \leadsto \ \ \) | \(\ds s \laptrans {\map {I_\R} t}\) | \(=\) | \(\ds \dfrac 1 s\) | |||||||||||
\(\ds \leadsto \ \ \) | \(\ds \laptrans {\map {I_\R} t}\) | \(=\) | \(\ds \dfrac 1 {s^2}\) |
$\blacksquare$
Sources
- 1965: Murray R. Spiegel: Theory and Problems of Laplace Transforms ... (previous) ... (next): Chapter $1$: The Laplace Transform: Laplace Transforms of some Elementary Functions: $2$.
- 1968: Murray R. Spiegel: Mathematical Handbook of Formulas and Tables ... (previous) ... (next): $\S 32$: Table of Special Laplace Transforms: $32.26$